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Castiglianos Theorem
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To Use This Method
You should have a some background with:
Deflection of a beam/cylinder due to:
Axial loading
Bending
Torsion
Calculating normal and polar moments of inertia.
Deriving equations for linear changes in quantities.
Using singularity functions (for more advanced applications;
no examples here explicitly show it, but it is often used inconjunction with Castiglianos Theorem.
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Contents
Definition4
Variations6
Examples10
Summary18
References19
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Definition
Determining the deflection of beams typically requires
repeated integration of singularity functions.
Castiglianos Theorem lets us use strain energies at the
locations of forces to determine the deflections.
The Theorem also allows for the determining of deflections for
objects with changing cross sectional areas.
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Definition
Castiglianos Theorem is given as:
Where is the deflection, U is the strain energy and P is the
force (or torque) at a certain point.
P
U
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Variations
Different loading conditions require different strain energies.
For axial loading:
L
Where P is the load, E is the materials Youngs Modulus
(usually either in GPa or ksi), A is the cross sectional area, and
L is the length.
L
EAdxPU
0
2
2
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Variations
For a material in bending:
L
Where M is the moment applied, and I is the area moment of
inertia.
L
EIdxMU
0
2
2
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Variations
For a material in torsion:
L
Where T is the torque applied, G is the Modulus of Rigidity,
and J is the polar moment of inertia.
L
GJ
dxTU0
2
2
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Variations
Note: except for the Youngs Modulus and Modulus of Rigidity
(E and G), it is not guaranteed that the other variables are not
functions of x.
Sometimes dimensions of the material change as functions ofx, and thus the moments of inertia change; and sometimes
the forces applied may vary with x.
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Examples
Imagine a cylinder attached to a fixed wall, with constant
diameter d=4 cm and length L=2 m, and a torque of 8 Nm is
applied. Assume G=120 GPa.
To find the displacement of the cylinder, we use Castiglianos
Theorem with the strain energy for torsion.
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Examples
Continuing example:
With
The area and moment of inertia are not changing, so we can
easily find the displacement.
L
GJ
dxT
T 0
2
2
4
32
dJ
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Examples
Continuing example:
Because differentiating and integrating are linear operations,
the partial derivative can be placed inside the integral:
.5305 mm
LLL
GJ
Tdxdx
GJ
T
TGJ
dxT
T 00
2
0
2
22
4904.
3210120
28
GJ
TL
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Examples
Imagine having a beam with a changing cross section shown
below, with an initial height of 3 m and a final height of 1 m,
with a constant base length of 2 m. The beam has a length of
6 m, with a Youngs Modulus of 120 GPa, and a force is applied
with magnitude P=10 kN.
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Examples
We will use Castiglianos Theorem applied for bending to solve
for the deflection where M is applied.
To find M, we need to consider the circumstances. At the wall
(x=0) the moment felt is the maximum moment or PL, but at
the end of the beam, the moment is zero because moments atthe locations do not contribute to the overall moments.
L
EI
dxM
P 0
2
2
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Examples
Continuing example:
And so:
The height is also a function of x, and the initial and final
heights can be used to formulate an equation:
PxPLxM )(
3
3
1)(
xhx
L
hhxh
i
if
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Examples
Continuing example:
And so the moment of inertia, as a function of x, is:
Substituting the functions we have derived into the equation
for the displacement:
33)3
3
1)(2(
12
1
12
1)( xbhxI
dx
xE
PxPL
P
L
03
2
)33
1(
6
12
)(
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Examples
Continuing example:
1.311 m
L
xE
dxxLPxPL
03)3
3
1(
6
1
))((
6
03)3
3
1(
6
1
)6)(1000060000(
xE
dxxx
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Summary
After viewing this tutorial, you should be confident in:
Identifying a situation (whether in axial loading, bending, or
torsion) where Castiglianos Theorem may be applied to solve
for the deflection in a beam or cylinder.
Generate equations for the changes in height, base, or even
force across the length of a beam/cylinder.
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References
Mechanics of MaterialsBeer/Johnson 5thEdition
Section 11.13 Deflections by Castiglianos Theorem
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